I need help in determining in these equations have solutions?

Question

m^2+m + 1 =0
p^2 -4p + 4=0
2x^2 +5x+5=0
4/3x^2-3x+3/4=0

Answer 1

there are solutions if
b^2-4ac>=0

m^2+m + 1 =0
1-4<0 no solution

p^2 -4p + 4=0
(p-2)^2=0
there is a solution. p1=p2=2

2x^2 +5x+5=0
a=2
b=5
c=5
25-40<0 no solution


4/3x^2-3x+3/4=0
a=4/3
b=-3
c=3/4
9-4*1=5>0 there is a solution

Answer 2

1)
m^2+m + 1 =0
m^2 + m + (1/2)^2 = -1 + (1/2)^2
(m+1/2)^2 = -4/4 + 1/4
(m+1/2)^2 = -3/4
sqrt [(m+1/2)^2] = sqrt (-3/4)
m + 1/2 = +- sqrt(3)/2 i
m = sqrt(3)/2 i - 1/2
or
m = - sqrt(3)/2 i - 1/2

2)
p^2 -4p + 4=0
p^2 -2p -2p + 4 =0
p(p-2) -2(p-2)=0
(p-2)(p-2)=0
(p-2)^2=0
p=2

3)
2x^2 +5x+5=0
x^2 + 5/2x + 5/2 =0
x^2 + 5/2x + (5/4)^2 = 25/16 - 5/2
(x+5/4)^2 = (25 - 40) / 16
sqrt [(x+5/4)^2] = sqrt (-15/16)
x+5/4 = +- sqrt(15)/4 i
x = sqrt(15)/4 i - 5/4
or
x = - sqrt(15)/4 i - 5/4

4)
4/3x^2-3x+3/4 =0 (3/4)
x^2-9/4x +9/16 =0
x^2 -9/4x + (9/8)^2 = 81/64 - 9/16
(x-9/8)^2 = (81-36)/64
(x-9/8)^2 = 45/64
sqrt [(x-9/8)^2] = sqrt(45/64)
x - 9/8 = +- sqrt(9*5)/8
x = [9 + 3 sqrt(5)] / 8
or
x= [9 - 3 sqrt(5)] / 8

Answer 3

every equation has got a solution.

It depends on whether it has got a real solution or not.

I feel u r searching for a method to find out whether these equations have got a real solution or not.

For an equation a X^2+b X + c = 0;

b^2 - 4*a*c is called the determinant (det).

The roots are real and distinct if b^2-4*a*c > 0.
The roots are real and equal if b^2-4*a*c = 0.
The roots are imaginary if b^2-4*a*c < 0.

So check the following in ur equations..

It says that det(equation 1) = -3 => complex roots.
det(equation 2) = 0 => real and equal roots (p=2).
det(equation 3) = -15 => complex roots.
det(equation 4) = 5 => roota are real and distinct
(x= 9/8 + (3/8)(5)^0.5 or x= 9/8 - (3/8)(5)^0.5.

Answer 4

m=-1+/-sqrt(-3)/2

p=2 double solution

x=-5+/-sqrt(-15)/4 can be written as = -5+/-i(sqrt15)

x=9/2+/-sqrt(72)